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1 Math 312, Fall 2012 Jerry L. Kazdan Problem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday. In addition to the problems below, you should also know how to solve the following problems from the text. Most are simple mental exercises. These are not to be handed in. Sec. 5.1, #28, 29, 31 Sec. 5.2 #33 1. [Bretscher, Sec. 5.1 #16] Consider the following vectors in R 4 u 1 =, u 2 =, u 3 =. Can you find a vector u 4 in R 4 such that the vectors u 1, u 2, u 3, u 4 are orthonormal? If so, how many such vectors are there? Solution: Since R 4 is four dimensional, you can extend these three orthonormal vectors to an orthonormal basis. First find a vector orthogonal to these three; then normalize it to be a unit vector. In this case, looking at the three given vectors, another immediately comes to mind: u 4 =, (1) with ˆ u := u 4 another possibility. These two are the only possibility since the orthogonal complement of the span of u 1, u 2, u 3 is one dimensional so a basis will have only one vector. After we have found one, which we call u 4, any other, say ˆ w must have the form w = c u 4 for some constant c. Because we want a unit vector, so c = ±1. 1 = w = c 2 u 4 = c 2, But what if my u 4 didn t immediately come to mind? Use the Gram-Schmidt process. Pick any vector not in the span of u 1, u 2, u 3. Almost any vector in R 4 will do. I will 1 try the simple w := 0 0. We want to write w in the form 0 w = a u 1 + b u 2 + c u 3 + z, 1

3 Solution: The orthogonality gives ab + d + fg = 0, ac + ed + f/2 = 0, bc + e + g/2 = 0. Because we want unit vectors, so we can t scale the second or third vectors, we need b = g = 0 and we can t simply let c = e = 0 (it took me a few minutes to grasp this). The orthogonality conditions are then d = 0, ac + f/2 = 0, e = 0. That these are unit vectors gives a 2 +f 2 = 1 and c 2 +1/4 = 1. Therefore c = ± 3/2, so f = ( 3)a, which in turn implies a = ± [Bretscher, Sec. 5.1 #26] Find the orthogonal projection P S of x := 49 into the subspace S of R 3 spanned by v 1 := 3 and v 2 := Solution: We are fortunate that the vectors v 1 and v 2 are orthogonal. We want to find constants a and b so that x = a v 1 + b v 2 + w, (2) where w is orthogonal to S. Then the desired projection will be P S x = a v 1 + b v 2. To find the scalars a and n, take the inner product of (2) with v 1 and then v 2 we find x, v 1 = a v 1 2 and x, v 2 = a v 2 2. Using the particular vectors in this problem, a = 11 and b = 1. Thus 19 P S x = 11 v 1 v 2 = [Bretscher, Sec. 5.1 #37] Consider a plane in R 3 with orthonormal basis u 1 and u 2. Let x be a vector in R 3. Find a formula for the reflection R x of x across the plane. Solution: The key is a picture (first try it in R 2 where is a line through the origin). Let P x be the orthogonal projection of x into the plane. Then w := P x = x P x is the projection of x orthogonal to. From the picture, to get the reflection, replace w by w 3

4 W=X P X X = P X + W W Thus, since x = P x + w, then P X R X = P X W R x = P x w = P x ( x P x) = 2P x x. In summary, orthogonal projections and reflections for a subspace are related by the simple formula R = 2P I. Note that the orthogonal projection, P x, is easy to compute if you know an orthonormal basis. All of this is very general. In this problem u 1 and u 2 are an orthonormal basis for the subspace, so P x = x, u 1 u 1 + x, u 2 u 2. Consequently, R x = 2( x, u 1 u 1 + x, u 2 u 2 ) x. 6. [Bretscher, Sec. 5.2 #32] Find an orthonormal basis for the plane x 1 +x 2 +x 3 = 0. Solution: Pick any point in the plane, say v 1 = (1, 1,0). This will be the first vector in our orthogonal basis. We use the Gram-Schmidt process to extend this to an orthogonal basis for the plane. Pick any other point in the plane, say w 1 := (1,0 1). Write it as w 1 = a v 1 + z, where z is perpendicular to v 1. Note that, although unknown, z will also be in the plane since it will be a linear combination of v 1 and w, both of which are in the plane. As usual, by taking the inner product of both sides of w 1 = a v 1 + z with v 1, we find Thus a = w 1, v 1 / v 1 2 = 1 2. z = w v 1 = 1 is in the plane and orthogonal to v 1. The vectors v 1 and z are an orthogonal basis for this plane. To get an orthonormal basis we just make these into unit vectors u 1 := v 1 v 1 = and u 2 := z 2 z = 1 0 3/2 1 4

5 7. Let be a linear space. A linear map P : is called a projection if P 2 = P (this P is not necessarily an orthogonal projection ). a) Show that the matrix P = ( ) is a projection. Draw a sketch of R2 showing the vectors (1,2), ( 1,0), (3,1) and (0,3) and their images under the map P. Also indicate both the image,, and kernel, W, of P. Solution: The image of P is the line x 1 = x 2 (the subspace ); the kernel is the x 1 axis (the subspace W ). See the figure below. b) Repeat this for the complementary projection Q := I P. Solution: The image of Q is the x 1 axis (the subspace W ); its kernel is the line x 1 = x 2 (the subspace ). P: Q: c) If the image and kernel of a projection P are orthogonal then P is called an orthogonal projection. [This of course now assumes that has an inner product.] Let M = ( 0 0 a c ). For which real value(s) of a and c is this a projection? An orthogonal projection? ( ) 0 ac Solution: M 2 = 0 c 2 so M 2 = M requires that ac = a and c 2 = c. The first requires that either a = 0 or c = 1. If a = 0 the second equation is satisfied if either c = 0 or c = 1. If a 0, then c = 1. Thus, the possibilities are: ( ) ( ) a P 1 :=, or P := (for any a). 0 1 For an orthogonal projection P, its image and kernal must be orthogonal. Since for P 1 its image is just 0, which is orthogonal to everything, it is an orthogonal projection. The kernel of P 2 is the horizontal axis. Its image consists of points of the form t(a, 1) for any scalar t. This straight line is perpendicular to the horizontal axis if (and only if) a = 0. Thus P 2 is an orthogonal projection if and only if a = 0. The remaining problems are from the Lecture notes on ectors kazdan/312f12/notes/vectors/vectors8.pdf 5 3 W

6 8. [p. 8 #5] The origin and the vectors X, Y, and X + Y define a parallelogram whose diagonals have length X + Y and X Y. Prove the parallelogram law X + Y 2 + X Y 2 = 2 X Y 2 ; This states that in a parallelogram, the sum of the squares of the lengths of the diagonals equals the sum of the squares of the four sides. Solution: The standard procedure is to express the norm in terms of the inner product and use the usual algebraic rules for the inner product. Thus X+Y 2 = X+Y, X+Y = X, X + X, Y + Y, X + Y, Y = X, X +2 X, Y + Y, Y, with a similar formula for X Y 2. After easy algebra, the result is clear. 9. [p. 8 #6] a) Find the distance D from the straight line 3x 4y = 10 to the origin. Solution: Note that the equation of the parallel line l through the origin is 3x 4y = 0, which we rewrite as N, X = 0, where N := (3, 4) and X = (x, y). Let X 0 be some point on the original line, so N, X 0 = 10. Then the desired distance D is the same as the distance from X 0 to the line l: N, X = 0, through the origin. But the equation for l says the vector N is perpendicular to the line l. Thus the distance D is the length of the projection of X 0 in the direction of N, that is, D = N, X 0 = 10 N 5 = 2. b) Find the distance D from the plane ax + by + cz = d to the origin (assume the vector N = (a,b,c) 0). Solution: The solution presented in the above special case generalizes immediately to give D = N, X 0 N = d N. 10. [p. 8 #8] a) If X and Y are real vectors, show that X, Y = 1 4 ( X + Y 2 X Y 2). (3) This formula is the simplest way to recover properties of the inner product from the norm. Solution: The straightforward procedure is the same as in Problem 8: rewrite the norms on the right in terms of the inner product and expand using algebra. 6

7 b) As an application, show that if a square matrix R has the property that it preserves length, so RX = X for every vector X, then it preserves the inner product, that is, RX, RY = X, Y for all vectors X and Y. Solution: We know that RZ = Z for any vector Z. This implies R(X + Y ) = X + Y for any vectors X and Y, and, similarly, R(X Y ) = X Y for any vectors X and Y. Consequently, by equation (3) (used twice) for all vectors X and Y. 4 RX, RY = R(X + Y ) 2 R(X Y ) 2 = X + Y 2 X Y 2 =4 X, Y 11. [p. 9 #10] a) If a certain matrix C satisfies X, CY = 0 for all vectors X and Y, show that C = 0. Solution: Since X can be any vector, let X = CY to show that CY 2 = CY, CY = 0. Thus CY = 0 for all Y so C = 0. b) If the matrices A and B satisfy X, AY = X, BY for all vectors X and Y, show that A = B. Solution: We have 0 = X, AY X, BY = X, (AY BY ) = X, (A B)Y for all X and Y so by part (a) with C := A B, we conclude that A = B. 12. [p. 9 #11 12] A matrix A is called anti-symmetric (or skew-symmetric) if A = A. a) Give an example of a 3 3 anti-symmetric matrix. Solution: The most general anti-symmetric 3 3 matrix has the form 0 a b a 0 c. b c 0 b) If A is any anti-symmetric matrix, show that X, AX = 0 for all vectors X. Solution: X, AX = A X, X = AX, X = X, AX. Thus 2 X, AX = 0 so X, AX = 0. 7

8 c) Say X(t) is a solution of the differential equation dx = AX, where A is an antisymmetric matrix. dt ( Show ) that X(t) = constant. [Remark: A special case is cos t that X(t) := satisfies X sin t = AX with A = ( ) so this problem gives another proof that cos 2 t + sin 2 t = 1]. Solution: Let E(t) := X(t) 2. We show that de/dt = 0. But, using part (b), de dt = d dt X(t), X(t) = 2 X(t), X (t) = 2 X(t), AX(t) = 0. 1-B This is a followup to problem 7. Bonus Problem [Please give this directly to Professor Kazdan] a) If a projection P is self-adjoint, so P = P, show that P is an orthogonal projection. b) Conversely, if P is an orthogonal projection, show that it is self-adjoint. [Last revised: November 10, 2012] 8

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